We find the y-intercept by evaluating [latex]f\left(0\right)[/latex].

[latex]\begin{array}{c}f\left(0\right)=3{\left(0\right)}^{2}+5\left(0\right)-2\hfill \\ \text{ }=-2\hfill \end{array}[/latex]

So the y-intercept is at [latex]\left(0,-2\right)[/latex].

For the x-intercepts, or roots, we find all solutions of [latex]f\left(x\right)=0[/latex].

[latex]0=3{x}^{2}+5x - 2[/latex]

In this case, the quadratic can be factored easily, providing the simplest method for solution.

[latex]0=\left(3x - 1\right)\left(x+2\right)[/latex]
[latex]\begin{array}{c}0=3x - 1\hfill & \hfill & \hfill & \hfill & 0=x+2\hfill \\ x=\frac{1}{3}\hfill & \hfill & \text{or}\hfill & \hfill & x=-2\hfill \end{array}[/latex]

So the roots are at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex]\left(-2,0\right)[/latex].

Analysis of the Solution

By graphing the function, we can confirm that the graph crosses the y-axis at [latex]\left(0,-2\right)[/latex]. We can also confirm that the graph crosses the x-axis at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex]\left(-2,0\right)[/latex].

We begin by solving for when the output will be zero.

[latex]0=2{x}^{2}+4x - 4[/latex]

Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.

[latex]f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex]

We know that a = 2. Then we solve for h and k.

[latex]\begin{array}{c}h=-\frac{b}{2a}\hfill & \hfill & \hfill & k=f\left(-1\right)\hfill \\ \text{ }=-\frac{4}{2\left(2\right)}\hfill & \hfill & \hfill & \text{ }=2{\left(-1\right)}^{2}+4\left(-1\right)-4\hfill \\ \text{ }=-1\hfill & \hfill & \hfill & \text{ }=-6\hfill \end{array}[/latex]

So now we can rewrite in standard form.

[latex]f\left(x\right)=2{\left(x+1\right)}^{2}-6[/latex]

We can now solve for when the output will be zero.

[latex]\begin{array}{c}0=2{\left(x+1\right)}^{2}-6\hfill \\ 6=2{\left(x+1\right)}^{2}\hfill \\ 3={\left(x+1\right)}^{2}\hfill \\ x+1=\pm \sqrt{3}\hfill \\ x=-1\pm \sqrt{3}\hfill \end{array}[/latex]

The graph has x-intercepts at [latex]\left(-1-\sqrt{3},0\right)[/latex] and [latex]\left(-1+\sqrt{3},0\right)[/latex].

Analysis of the Solution

We can check our work by graphing the given function on a graphing utility and observing the roots.

Let’s begin by writing the quadratic formula: [latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex].

When applying the quadratic formula, we identify the coefficients a, b, and c. For the equation [latex]{x}^{2}+x+2=0[/latex], we have a = 1, b = 1, and c = 2. Substituting these values into the formula we have:

[latex]\begin{array}{c}x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \text{ }=\frac{-1\pm \sqrt{{1}^{2}-4\cdot 1\cdot \left(2\right)}}{2\cdot 1}\hfill \\ \text{ }=\frac{-1\pm \sqrt{1 - 8}}{2}\hfill \\ \text{ }=\frac{-1\pm \sqrt{-7}}{2}\hfill \\ \text{ }=\frac{-1\pm i\sqrt{7}}{2}\hfill \end{array}[/latex]

The solutions to the equation are [latex]x=\frac{-1+i\sqrt{7}}{2}[/latex] and [latex]x=\frac{-1-i\sqrt{7}}{2}[/latex] or [latex]x=\frac{-1}{2}+\frac{i\sqrt{7}}{2}[/latex] and [latex]x=\frac{-1}{2}-\frac{i\sqrt{7}}{2}[/latex].

a. The ball reaches the maximum height at the vertex of the parabola.

[latex]\begin{array}{c} h=-\frac{80}{2\left(-16\right)} \text{ }=\frac{80}{32}\hfill \\ \text{ }=\frac{5}{2}\hfill \\ \text{ }=2.5\hfill \end{array}[/latex]

The ball reaches a maximum height after 2.5 seconds.

b. To find the maximum height, find the y coordinate of the vertex of the parabola.

[latex]\begin{array}{c}k=H\left(-\frac{b}{2a}\right)\hfill \\ \text{ }=H\left(2.5\right)\hfill \\ \text{ }=-16{\left(2.5\right)}^{2}+80\left(2.5\right)+40\hfill \\ \text{ }=140\hfill \end{array}[/latex]

The ball reaches a maximum height of 140 feet.

c. To find when the ball hits the ground, we need to determine when the height is zero, [latex]H\left(t\right)=0[/latex].

We use the quadratic formula.

[latex]\begin{array}{c} t=\frac{-80\pm \sqrt{{80}^{2}-4\left(-16\right)\left(40\right)}}{2\left(-16\right)}\hfill \\ \text{ }=\frac{-80\pm \sqrt{8960}}{-32}\hfill \end{array}[/latex]

Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

[latex]\begin{array}{c}t=\frac{-80-\sqrt{8960}}{-32}\approx 5.458\hfill & \text{or}\hfill & t=\frac{-80+\sqrt{8960}}{-32}\approx -0.458\hfill \end{array}[/latex]

The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds.